# Coin on a turntable problem in a noninertial frame

• Natchanon
In summary, this conversation discusses the problem of a coin placed on a sticky turntable rotating with constant angular acceleration. The goal is to find the three fictitious forces acting on the coin and determine the time elapsed before the coin begins to slide, given the friction coefficient. The net force keeping the coin in circular motion must be zero, and the equation for static friction is |Fstatic friction|≤μs|N|.
Natchanon
1. The problem statement
A small coin of mass m is placed on a turntable a distance a from the axis. Imagine that the surface is a little sticky, so the coin does not slide immediately. The turntable (initially at rest) is rotated at constant angular acceleration α, beginning at t=0.
a.) Find the three fictitious forces as functions of time. (find both magnitudes and directions in polar unit vectors)
b.) If you are given the static friction coefficient μ, find the time elapsed before the coin begins to slide. (Hint: This will depend on the net lateral force on the co in.)

## Homework Equations

ma_rot = f - mω x (ω x r) - 2mω x v -mα x r

## The Attempt at a Solution

I want to know if I understand this correctly. The coin will start to slide when the net force keeping it in circular motion, that is friction + coriolis + centrifugal + Euler, equals zero, correct? Does that mean all I need to do is to find the net force from its x and y components and set it equal to zero?

Natchanon said:
The coin will start to slide when the net force keeping it in circular motion, that is friction + coriolis + centrifugal + Euler, equals zero, correct?
No, that's wrong in a couple of ways.

If the turntable were stationary, the net force on the coin would be zero, but it won't slide.
What is the equation relating the force of static friction to its coefficient? Be careful, it is more complicated than for kinetic.

Secondly, it is not only the radial forces that need to be considered. If a car brakes on a bend it is more likely to skid - why?

Natchanon
haruspex said:
No, that's wrong in a couple of ways.

If the turntable were stationary, the net force on the coin would be zero, but it won't slide.
What is the equation relating the force of static friction to its coefficient? Be careful, it is more complicated than for kinetic.

Secondly, it is not only the radial forces that need to be considered. If a car brakes on a bend it is more likely to skid - why?

friction force = μmg. If a car brakes on a bend it is more likely to skid because your angular velocity increases in magnitude, so the centrifugal force becomes so large that the friction between the tires and road can't keep you on the road.

Natchanon said:
friction force = μmg
No, as I wrote, it is more complicated than that.
E.g. what is the frictional force on a block sitting on a level floor with nothing pushing from the side?
Natchanon said:
If a car brakes on a bend it is more likely to skid because your angular velocity increases in magnitude
No, the angular velocity would start to reduce, yet immediately it wouid be more likely to skid. What has changed in terms of linear acceleration?

haruspex said:
No, as I wrote, it is more complicated than that.
E.g. what is the frictional force on a block sitting on a level floor with nothing pushing from the side?

No, the angular velocity would start to reduce, yet immediately it wouid be more likely to skid. What has changed in terms of linear acceleration?

It increases to the opposite direction of the linear velocity.

Natchanon said:
It increases to the opposite direction of the linear velocity.
Right, so what happens to the magnitude of the net acceleration?

haruspex said:
Right, so what happens to the magnitude of the net acceleration?
It decreases.

haruspex said:
Right, so what happens to the magnitude of the net acceleration?
I mean "increases".

Natchanon said:
I mean "increases".
Right, and that increases the demand for static friction if skidding is to be avoided.
This is what the hint was for: the static friction has to supply the net acceleration, which includes both a radial and a tangential component (in the lab frame).
Indeed, you did list the Euler force, so you had that covered, but you referred to the list as the forces "required to keep it in circular motion". Only the radial force is needed for that.

On the other matter, the correct equation for the force of static friction is |Fstatic friction|≤μs|N|. Thus, the net of the forces you listed will be zero right up until the point where slipping occurs.

Natchanon
haruspex said:
Right, and that increases the demand for static friction if skidding is to be avoided.
This is what the hint was for: the static friction has to supply the net acceleration, which includes both a radial and a tangential component (in the lab frame).
Indeed, you did list the Euler force, so you had that covered, but you referred to the list as the forces "required to keep it in circular motion". Only the radial force is needed for that.

On the other matter, the correct equation for the force of static friction is |Fstatic friction|≤μs|N|. Thus, the net of the forces you listed will be zero right up until the point where slipping occurs.
So that means the coin slides when Euler force + Centrifugal force = Frictional force.

Natchanon said:
So that means the coin slides when Euler force + Centrifugal force = Frictional force.
No, you are still missing my point about static friction. That equality will hold true at all times that the coin is not sliding.
Also, it should take the form ΣF=0: When on the point of sliding, Euler force + Centrifugal force + Maximum static Frictional force=0.

haruspex said:
No, you are still missing my point about static friction. That equality will hold true at all times that the coin is not sliding.
Also, it should take the form ΣF=0: When on the point of sliding, Euler force + Centrifugal force + Maximum static Frictional force=0.
Oh, I see. So, not frictional force, but the maximum static frictional force which is the threshold. I forgot the minus sign on the right side.

Natchanon said:
Oh, I see. So, not frictional force, but the maximum static frictional force which is the threshold. I forgot the minus sign on the right side.
Yes.

Natchanon

## 1. What is the "Coin on a turntable problem in a noninertial frame"?

The "Coin on a turntable problem in a noninertial frame" is a physics problem that explores the behavior of an object, such as a coin, placed on a turntable that is rotating at a constant angular velocity. The problem is typically studied in a noninertial frame of reference, meaning that the frame is accelerating or rotating with respect to an inertial frame.

## 2. What is the significance of studying this problem?

Studying the "Coin on a turntable problem in a noninertial frame" allows us to better understand the effects of noninertial frames on the motion of objects. This can have practical applications in fields such as engineering, where understanding the behavior of objects in rotating systems is important.

## 3. How is this problem solved?

This problem can be solved using the laws of physics, particularly Newton's laws of motion and the concepts of centripetal force and centrifugal force. The solution involves analyzing the forces acting on the object, such as the normal force and the frictional force, and applying Newton's second law of motion.

One common misconception is that the coin will fly off the turntable due to centrifugal force. In reality, the coin will remain on the turntable due to the presence of a normal force, which acts perpendicular to the surface of the turntable and prevents the coin from flying off.

Another misconception is that the coin experiences a centrifugal force. In reality, the coin experiences a centripetal force, which is directed towards the center of the turntable and keeps the coin in circular motion.

## 5. Are there any real-life examples of this problem?

Yes, there are many real-life examples of this problem, such as a coin or a cup placed on a rotating platform, or objects on a merry-go-round. The principles and equations used to solve this problem can also be applied to more complex rotating systems, such as rotating satellites or amusement park rides.

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